Similar triangles and polygons worksheet help with number 1

You ask 125 randomly chosen students their favorite sport. There are 1300 students in the school. Based on the results of the su

yan [13]

780 students

Set up a proportion.

75 / 125 = x/1300

125 10.4 = 1300

75 * 10.4 = 780

x=780

5 0

3 years ago

Complete the puzzle by clicking on the puzzle piece that contains equivalent expression.

Liono4ka [1.6K]

Answer:

-5 when ...

Step-by-step explanation:

The rules of exponents can help you simplify the given product.

<h3>Rules</h3>

(a/b)^c = (a^c)/(b^c)

(ab)^c = (a^c)(b^c)

(a^b)(a^c) = a^(b+c)

(a^b)/(a^c) = a^(b-c)

<h3>Application</h3>

$\left(\dfrac{x}{y}\right)^2(-x^2y)^3=\\\\\dfrac{x^2(-1)^3(x^2)^3y^3}{y^2}=-x^{2+2\cdot3}y^{3-2}=-x^8y$

This expression does not match any of those offered.

When x=-1 and y=5, this becomes ...

$-(-1)^8(5)=\boxed{-5}$

8 0

2 years ago

"A study conducted at a certain college shows that 56% of the school's graduates find a job in their chosen field within a year

KiRa [710]

Answer:

99.27% probability that among 6 randomly selected graduates, at least one finds a job in his or her chosen field within a year of graduating.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they find a job in their chosen field within one year of graduating, or they do not. The probability of a student finding a job in their chosen field within one year of graduating is independent of other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

56% of the school's graduates find a job in their chosen field within a year after graduation.

This means that $p = 0.56$

Find the probability that among 6 randomly selected graduates, at least one finds a job in his or her chosen field within a year of graduating.

This is $P(X \geq 1)$ when n = 6.

Either none find a job, or at least one does. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.(0.56)^{0}.(0.44)^{6} = 0.0073$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0073 = 0.9927$

99.27% probability that among 6 randomly selected graduates, at least one finds a job in his or her chosen field within a year of graduating.

8 0

3 years ago

Please help me solve this, and show your work

zlopas [31]

$6r+5=11r\ \ \ \ |\text{subtract 6r from both sides}\\\\5=5r\ \ \ \ |\text{divide both sides by 5}\\\\\boxed{r=1}\\-------------------------\\\dfrac{2x-4}{2}=6x+3\\\\\dfrac{2x}{2}-\dfrac{4}{2}=6x+3\\\\x-2=6x+3\ \ \ \ \ |\text{add 2 to both sides}\\\\x=6x+5\ \ \ \ |\text{subtract 6x from both sides}\\\\-5x=5\ \ \ \ \ |\text{divide both sides by -5}\\\\\boxed{x=-1}$

6 0

3 years ago

HELP A.S.A.P!!!! The data to represent average test scores for a class of 16 students includes an outlier value of 78. If the ou

kolezko [41]

When the outlier is removed The mean would increase.

Step-by-step explanation:

Outliers are the value that lies outside the set of data. i.e. the data doesn't fit in the group.

Removing outliers can affect mean more than median. Because outlier removal sometimes affect the median but in a small percentage or sometimes it will remain the same.

Let us prove, consider 16 values one with 78 and other 15 with values nearly 85, 84 that will give the mean value as 84.

78, 84, 82, 86, 84, 85, 83, 86, 82, 85, 84, 83, 86, 85, 86, 85.

Mean = $\frac{Sum of the values}{Number of values}$ .